The value of $$\sum_{r=1}^{30} \frac{r^{2} \left(^{30}C_{r}\right)^{2}}{^{30}C_{r-1}}$$ is $$\alpha \times 2^{29}$$ then $$\alpha$$ is equal to

Let's first simplify the ratio of the combinations $$\frac{^{30}C_{r}}{^{30}C_{r-1}}$$.
Using the identity $$\frac{^{n}C_{r}}{^{n}C_{r-1}} = \frac{n-r+1}{r}$$, we can substitute $$n = 30$$:

$$\frac{^{30}C_{r}}{^{30}C_{r-1}} = \frac{30-r+1}{r} = \frac{31-r}{r}$$

Now, substitute this back into our general term inside the summation:

$$\frac{r^{2} \left(^{30}C_{r}\right)^{2}}{^{30}C_{r-1}} = r^2 \cdot \left(^{30}C_{r}\right) \cdot \left(\frac{^{30}C_{r}}{^{30}C_{r-1}}\right)$$ $$= r^2 \cdot \left(^{30}C_{r}\right) \cdot \left(\frac{31-r}{r}\right)$$ $$= r(31-r) \cdot ^{30}C_{r}$$ $$= (31r - r^2) \cdot ^{30}C_{r}$$

$$\sum_{r=1}^{30} (31r - r^2) \cdot ^{30}C_{r} = 31 \sum_{r=1}^{30} r \cdot ^{30}C_{r} - \sum_{r=1}^{30} r^2 \cdot ^{30}C_{r}$$

1. First Sum:
   $$\sum_{r=1}^{n} r \cdot ^{n}C_{r} = n \cdot 2^{n-1}$$

   For $$n = 30$$:
   $$\sum_{r=1}^{30} r \cdot ^{30}C_{r} = 30 \cdot 2^{29}$$

2. Second Sum:
   $$\sum_{r=1}^{n} r^2 \cdot ^{n}C_{r} = n(n+1) \cdot 2^{n-2}$$

   For $$n = 30$$:
   $$\sum_{r=1}^{30} r^2 \cdot ^{30}C_{r} = 30(31) \cdot 2^{28} = 30 \cdot 31 \cdot \frac{2^{29}}{2} = 15 \cdot 31 \cdot 2^{29}$$

Substitute these values back into our equation:

$$\text{Sum} = 31 \left(30 \cdot 2^{29}\right) - \left(15 \cdot 31 \cdot 2^{29}\right)$$
Factor out $$31 \cdot 2^{29}$$:

$$\text{Sum} = 31 \cdot 2^{29} \cdot (30 - 15)$$ $$\text{Sum} = 31 \cdot 2^{29} \cdot 15$$ $$\text{Sum} = (31 \times 15) \times 2^{29}$$ $$\text{Sum} = 465 \times 2^{29}$$

Comparing this with the given right-hand side ($$\alpha \times 2^{29}$$):
$$\alpha = 465$$